Question Number 25773 by Ali Makhzoum last updated on 14/Dec/17
![The least value of the function φ(x) = ∫_(5π/4) ^x (3 sin t+4 cos t) dt on the interval [ ((5π)/4), ((4π)/3) ] is](Q25773.png)
$$\mathrm{The}\:\mathrm{least}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function} \\ $$$$\:\:\:\phi\left({x}\right)\:=\:\underset{\mathrm{5}\pi/\mathrm{4}} {\overset{{x}} {\int}}\:\left(\mathrm{3}\:\mathrm{sin}\:{t}+\mathrm{4}\:\mathrm{cos}\:{t}\right)\:{dt} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{interval}\:\left[\:\frac{\mathrm{5}\pi}{\mathrm{4}},\:\frac{\mathrm{4}\pi}{\mathrm{3}}\:\right]\:\mathrm{is} \\ $$
Answered by ajfour last updated on 14/Dec/17
![φ(x)=(4sin t−3cos t)∣_(5π/4) ^x =4sin x−3cos x−(−(4/(√2))+(3/(√2))) =4sin x−3cos x+(1/(√2)) =5sin (x−α)+(1/(√2)) where α=sin^(−1) ((4/5)) ((d[φ(x)])/dx) =5cos (x−α) if ((5π)/4) ≤ x ≤ ((4π)/3) 225°−53° ≤ x−α ≤ 240°−53° ⇒ 172° ≤ x−α ≤ 187° so ((d[φ(x)])/dx) =5cos (x−α) < 0 function φ(x) is decreasing in this interval; hence in this interval φ(x) is least for x=4π/3 φ(((4π)/3))=4sin (((4π)/3))−3cos (((4π)/3))+(1/(√2)) =4×(−((√3)/2))−3(−(1/2))+(1/(√2)) =(3/2)+(1/(√2))−2(√3) .](Q25775.png)
$$\phi\left({x}\right)=\left(\mathrm{4sin}\:{t}−\mathrm{3cos}\:{t}\right)\mid_{\mathrm{5}\pi/\mathrm{4}} ^{{x}} \\ $$$$\:\:=\mathrm{4sin}\:{x}−\mathrm{3cos}\:{x}−\left(−\frac{\mathrm{4}}{\sqrt{\mathrm{2}}}+\frac{\mathrm{3}}{\sqrt{\mathrm{2}}}\right) \\ $$$$\:\:=\mathrm{4sin}\:{x}−\mathrm{3cos}\:{x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\:\:=\mathrm{5sin}\:\left({x}−\alpha\right)+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{where}\:\alpha=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\:\frac{{d}\left[\phi\left({x}\right)\right]}{{dx}}\:=\mathrm{5cos}\:\left({x}−\alpha\right) \\ $$$${if}\:\:\:\:\:\:\:\frac{\mathrm{5}\pi}{\mathrm{4}}\:\leqslant\:{x}\:\leqslant\:\frac{\mathrm{4}\pi}{\mathrm{3}} \\ $$$$\:\:\mathrm{225}°−\mathrm{53}°\:\leqslant\:{x}−\alpha\:\leqslant\:\mathrm{240}°−\mathrm{53}° \\ $$$$\Rightarrow\:\:\:\:\:\:\:\mathrm{172}°\:\leqslant\:{x}−\alpha\:\leqslant\:\mathrm{187}° \\ $$$$\:\:{so}\:\frac{{d}\left[\phi\left({x}\right)\right]}{{dx}}\:=\mathrm{5cos}\:\left({x}−\alpha\right)\:<\:\mathrm{0} \\ $$$${function}\:\phi\left({x}\right)\:{is}\:{decreasing}\:{in} \\ $$$${this}\:{interval};\:{hence}\:{in}\:{this} \\ $$$${interval}\:\phi\left({x}\right)\:{is}\:{least}\:{for}\:{x}=\mathrm{4}\pi/\mathrm{3} \\ $$$$\phi\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)=\mathrm{4sin}\:\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)−\mathrm{3cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}×\left(−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\mathrm{2}\sqrt{\mathrm{3}}\:. \\ $$